= Solution
Fix $x$ with $\operatorname{dist}(x,\partial\Omega)>\sigma$ and use $\eta_x(y)=\phi_\sigma(x-y)$ as the <test function> in part (a). Its support lies compactly inside $\Omega$. Differentiating the <convolution> under the integral is allowed, and $D_{y_i}\eta_x=-D_{x_i}\phi_\sigma(x-y)$ while $\Delta_y\eta_x=\Delta_x\phi_\sigma(x-y)$. Consequently
$$
\begin{aligned}
\Delta u_\sigma(x)&=\int_\Omega u(y)\Delta_y\eta_x(y)\,dy\\
&=-\sum_i\int_\Omega b^i(y)u(y)D_{y_i}\eta_x(y)\,dy+\int_\Omega(cu+f)(y)\eta_x(y)\,dy\\
&=\sum_iD_{x_i}\int_\Omega b^i(y)u(y)\phi_\sigma(x-y)\,dy+(cu)_\sigma(x)+f_\sigma(x).
\end{aligned}
$$
Thus
$$
\boxed{\Delta u_\sigma=\sum_iD_i(b^iu)_\sigma+(cu)_\sigma+f_\sigma}
$$
at every stated interior point. This is the <convolution derivative identity>. The <convolutions> are local, so no integrability of the smooth coefficients all the way to the boundary is needed. In particular, $(b^iu)_\sigma$ is the <convolution> of the product; it must not be replaced by $b^iu_\sigma$ for a variable coefficient.
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